# Conservation of angular momentum about CM

So, recently while I was revisiting rotational dynamics, I came across this question: A rod AB of length L stands vertically on a horizontal floor and its end B is at height L. Now the end B starts falling down because of some slight disturbance and end A does not slip. Calculate the speed with which the end B hits the ground.

See, this question can be easily done by applying mechanical energy conservation. Since the end A is not moving there must be enough friction there, even then we can easily apply energy conservation because the point where friction acts (end A) does not displace, hence work done by friction is zero. But why can't we apply angular momentum conservation about the centre of mass of that rod as there won't be any net torque about CM? (As I said, the point where friction acts doesn't move so its work is zero, and in case of gravity, the torque due to weight will be zero as we are considering our axis to pass through CM.)

• $PE = m g L/2; KE = \frac12 m v^2 + \frac12 I \omega^2;$ – Narasimham Nov 26 '19 at 18:31
• The normal force at the rod contact with the ground produces a torque about the cm. – Bill Watts Nov 26 '19 at 18:35
• @Narasimham Thanks but I am asking why can't we solve it by using angular conservation as friction does no work – jatin Goyal Nov 26 '19 at 18:35
• Torque is $r\times F$, so the friction force produces a torque also even if it does no work. But if your cm is your relative stationary point, the contact with the floor is moving relative to that. – Bill Watts Nov 26 '19 at 18:43
• @BillWatts Thanks sir, but i have some query related to this.. 1) if we have a rod and a torque acts on it at some random point A , now if we consider another point B on that same rod so can we apply angular momentum conservation given that the point A is at rest throughout from B's frame of reference – jatin Goyal Nov 26 '19 at 18:47